# Showing commutativity of the ground monoid in a monoidal category

I have been trying to understand why the ground monoid in a monoidal category is commutative and every proof I have seen essentially uses the same thing to prove it namely by using the fact(apparently) that if every other subdiagram of a diagram is commutative then the remaining subdiagram is commutative. They all leave the proof of this fact and I don't understand why this should hold. I know that it has something to do with the fact that all the arrows in this diagram are isomorphisms but I don't know the whole picture.

I have added this picture from Vladimir's book on monoidal categories the emphasis here is on the 2nd last paragraph where it states "consequently the lower right triangle commutes" I don't understand this point.

• Yes, you need the fact that all the arrows are isomorphisms. The point is that, if $f$ and $h$ are isomorphisms, then $f \circ g_1 \circ h = f \circ g_2 \circ h$ if and only if $g_1 = g_2$, so you can pre-compose and post-compose as many isomorphisms as you want to prove the identity you need. Commented Sep 8, 2020 at 3:40
• That said, in my view, the coherence isomorphisms are actually not the most important part of the proof. The claim is true in any monoidal category – so in particular it is true in any strict monoidal category, where all the coherence isomorphisms are identity morphisms; yet even in this case there is still something to be proved, so the heart of the proof isn't equation (1.4) on that page. (Spoiler: it is the Eckmann–Hilton argument.) Commented Sep 8, 2020 at 3:46
• @Zhen Lin I understand that part(i.e I get how left tensoring with I implies the equations without I left tensored) but I still don't understand why the lower right triangle must commute. Commented Sep 8, 2020 at 3:47
• It's the same reasoning, except $f$ and $h$ are more complicated and there are more intermediate steps. Start by taking $g_1 = \textrm{id}_I \otimes (l_X \otimes \textrm{id}_Y)$ and $g_2 = (\textrm{id}_I \otimes l_{X \otimes Y}) \circ (\textrm{id}_I \otimes a_{I, X, Y})$. Commented Sep 8, 2020 at 3:53
• @Zhen Lin Thanks Now I understand what you were talking about. I will give this a try now Commented Sep 8, 2020 at 3:59

$$\text{End}(1)$$ has two monoid operations. One is composition and the other is tensor product, by which I mean: if $$f, g : 1 \to 1$$ are two morphisms they have a tensor product $$f \otimes g : 1 \otimes 1 \to 1 \otimes 1$$, and then we use the unit maps $$1 \otimes 1 \cong 1$$ to turn this into another endomorphism of $$1$$. These two monoid operations satisfy the interchange law and so you can run the Eckmann-Hilton argument on them: see for example this blog post.
• I proved it the way you described but I had to use this identity $l_{I} =r_{I}$ which is also something I need to prove. Commented Sep 8, 2020 at 7:31